Note n271s n applies m 00 2mcosfmsino and eq

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Unformatted text preview: nd b such that must be with the at R, consistent idea as solenoid. [Note: N/271"s = n applies (m. 0)0 = 2mcos()f+msin()O, and Eq. 5.87 <=> Eq. 5.86. Qed 3(m.f)f - m = 3m cos() f - mcos()f+msin()O s is essentially fLoI 2 (a)m=Ia=lhrR2i.1 in circumference, so that if the toroid is large for s ~ R; ( constant over the cross-section.] 47rR2 S - R2) Z, 2In(R/b)= 1- (b/R)2. I'll use a = b = R. Then A = (b) B 5.18 Problem ~ I~; I :~2 (2 cos0 f + sin 0 8) . . fLoI for s { 1.6.2, In(s/ R) Z, - 21!' J . da is independent ;::: R. } It doesn't matter. According to Theorem 2, in Sect. of surface, for any given boundary line, provided0 that J = z, f = i (for z > 0), so B ~for steadyi currents< (Eq. = 11", = -i, so the field Problem 5.26 z axis, = 0, r is divergenceless, which it is, (c) On the (for z 0, 0 5.31).f Problem 5.19 is the = Kx ~ B = xfLoK Y (plus for The 0, minus for z (Eq. 5.38) reduces (for zz » R) to B ~ /LoIR2/2IzI3, K same, with Izl3 in2place of Z3). z < exact answer &...
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